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Locus in Space 7. The locus of the vertex of a right triangle having a given hypotenuse is the sphere having the hypotenuse as its diameter.

Fig. 7-33 For t < 0, the 50-V source results in inductor current 50=20 2:5 A. The 5-A current source is applied for t > 0. As t ! 1, this current divides equally between the two 10-

L2 _

and so, with t in ms and using iL 0 iL 0 2:5 A, iL iL 0 iL 1 e t= iL 1 5:0e 100t 2:5 A

The analytic (or coordinate) geometry of two dimensions can easily be extended to three dimensions. Figure 17-13 shows the two- and three-dimensional axes; the z-axis is perpendicular to both the x- and y-axes. In the figure, arrows indicate the positive directions, and dashed lines indicate the negative axes. Four specific extensions from geometry in a plane to geometry in space follow.

Finally, using unit step functions to combine the expressions for t < 0 and t > 0, iL 2:5u t 5:0e 100t 2:5 u t A

Fig. 17-13

L2 _

The switch in Fig. 7-34 has been in position 1 for a long time; it is moved to 2 at t 0. the expression for i, for t > 0.

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Coordinate Geometry in a Plane 1. Coordinates: P1 (x1, y1), P2(x2, y2) 2. Midpoint of P1P2: x1 x2 y1 y2 , yM xM 2 2 3. Distance P1P2: d 2(x2 x1)2 (y2 y1)2 4. Equation of a circle having the origin as center and radius r: x2 y2 r2

Fig. 7-25 For the ampli er circuit of Example 7.4, reduce the value of the bypass capacitor CS to 0:01 F so that RS no longer appears shorted to ac signals and assess the impact on voltage gain. (Netlist code available from author website.) Ans: Av 1:22 1398 The series-connected JFETs of Fig. 4-23 are identical, with 70; rds 30 k ; RG 100 k ; and RD RL 4 k . Find (a) the voltage-gain ratio Av vL =vi ; b the current-gain ratio Ai iL =ii , and (c) the output impedance Ro . Ans: a Av 9:32; b Ai 233; c Ro 2:16 M The JFET ampli er of Fig. 4-33 has RG 1 M ; RS1 800 ; RS2 1:2 k ; and RL 1 k . The JFET obeys (4.2) and is characterized by IDSS 10 mA, Vp0 4 V; VGSQ 2 V; and 60. Determine (a) gm by use of (7.3), (b) rds , and (c) the voltage-gain ratio Av vL =vi . Ans: a 2:5 mS; b 24 k ; c 0:52

With the switch on 1, i 0 50=40 1:25 A. With an inductance in the circuit, i 0 i 0 . after the switch has been moved to 2, i 1 10=40 0:25 A. In the above notation, B i 1 0:25 A and the time constant is L=R 1=2000 s. A i 0 B 1:00 A Then, for t > 0, A

Coordinate Geometry in Space 1. Coordinates: P1 (x1, y1, z1), P2 (x2, y2, z2) 2. Midpoint of P1P2: x1 x2 y1 y2 z1 z2 , yM , zM xM 2 2 2 3. Distance P1P2: d 2(x2 x1)2 (y2 y1)2 (z2 z1)2 4. Equation of a sphere having the origin as center and radius r: x2 y2 z2 r2

i 1:00e 2000t 0:25

CHAP. 7]

The area of each face of the cube in Fig. 17-14 is A S e2. The total surface area S of the cube is then 6e2

CHAP. 7]

The areas of the six rectangles that make up the rectangular solid in Fig 17-15 are A A A lw lh for top and bottom faces for front and back faces

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